At 298 K, nitrous acid (HNO2) dissociates in water with a Ka of 0.00071.
a)Calculate G° for the dissociation of HNO2. (Use Ka to calculate the answer.)
b) Calculate G if [H3O+] = 0.00070 M, [NO2 - ] = 0.16 M, and [HNO2 ] = 0.21 M.
DGo = -RTlnK, then
DG = DGo+RTlnK where you substitute concns for those materials making up Ka.
a) To calculate the standard Gibbs free energy (ΔG°) for the dissociation of HNO2, we need to use the equation:
ΔG° = -RT ln(Ka)
Where:
R = Gas constant (8.314 J/mol·K)
T = Temperature in Kelvin
Ka = Acid dissociation constant
Given:
Temperature, T = 298 K
Ka = 0.00071
Substituting these values into the equation, we have:
ΔG° = - (8.314 J/mol·K) * (298 K) * ln(0.00071)
Calculating this expression using a calculator, we get:
ΔG° = - 5904 J/mol
Therefore, ΔG° for the dissociation of HNO2 at 298 K is -5904 J/mol.
b) To calculate the Gibbs free energy (ΔG) for the given concentrations of [H3O+], [NO2-], and [HNO2], we can use the equation:
ΔG = ΔG° + RT ln(Q)
Where:
ΔG° = Standard Gibbs free energy (calculated in part a)
R = Gas constant (8.314 J/mol·K)
T = Temperature in Kelvin
Q = Reaction quotient, calculated as [NO2-] / ([H3O+] * [HNO2])
Given:
ΔG° = -5904 J/mol
Temperature, T = 298 K
[H3O+] = 0.00070 M
[NO2-] = 0.16 M
[HNO2] = 0.21 M
Calculating the reaction quotient, Q:
Q = (0.16) / ((0.00070) * (0.21))
Q ≈ 1023.80
Substituting all the values into the equation, we have:
ΔG = -5904 J/mol + (8.314 J/mol·K) * (298 K) * ln(1023.80)
Calculating this expression using a calculator, we get:
ΔG ≈ -4783 J/mol
Therefore, ΔG for the given concentrations is approximately -4783 J/mol.
To calculate the standard free energy change (ΔG°) for the dissociation of nitrous acid (HNO2) in water, you can use the equation:
ΔG° = -RT ln(Ka)
where:
ΔG° = standard free energy change
R = gas constant (= 8.314 J/(mol·K))
T = temperature in Kelvin (298 K)
ln = natural logarithm
Ka = acid dissociation constant
a) Calculate G° for the dissociation of HNO2:
Plug in the values into the equation:
ΔG° = - (8.314 J/(mol·K)) * (298 K) * ln(0.00071)
Simplify and calculate:
ΔG° ≈ - (8.314 J/(mol·K)) * (298 K) * (-7.253)
ΔG° ≈ 17,282.4 J/mol
b) To calculate ΔG when given the concentrations of [H3O+], [NO2-], and [HNO2], you can use the equation:
ΔG = ΔG° + RT ln(Q)
where:
ΔG = free energy change
ΔG° = standard free energy change (calculated in part a)
R = gas constant (= 8.314 J/(mol·K))
T = temperature in Kelvin (298 K)
ln = natural logarithm
Q = reaction quotient
The reaction quotient (Q) is calculated using the concentrations of the products and reactants:
Q = ([H3O+] * [NO2-]) / [HNO2]
Plug in the values into the equation:
Q = (0.00070 M * 0.16 M) / 0.21 M
Simplify and calculate:
Q ≈ 0.3373
Now, plug in the values of ΔG°, R, T, and Q into the equation:
ΔG = 17,282.4 J/mol + (8.314 J/(mol·K)) * (298 K) * ln(0.3373)
Simplify and calculate:
ΔG ≈ 17,282.4 J/mol + (8.314 J/(mol·K)) * (298 K) * (-1.087)
ΔG ≈ -8,350.3 J/mol